MHT CET · Maths · Three Dimensional Geometry
If a plane meets the axes \(\mathrm{X}, \mathrm{Y}, \mathrm{Z}\) in \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) respectively such that centroid of \(\triangle \mathrm{ABC}\) is \((1,2,3)\), then the equation of the plane is
- A \(x+2 y+3 z=1\)
- B \(x+\frac{y}{2}+\frac{z}{3}=3\)
- C \(\frac{x}{3}+\frac{y}{6}+\frac{z}{9}=1\)
- D \(\frac{x}{4}+\frac{y}{8}+\frac{z}{12}=1\)
Answer & Solution
Correct Answer
(C) \(\frac{x}{3}+\frac{y}{6}+\frac{z}{9}=1\)
Step-by-step Solution
Detailed explanation
Let \(A=(x, 0,0) ; B=(0, y, 0) ; C=(0,0, z)\) Centroid of \(\triangle A B C\) is \((1,2,3)\)
\(
\therefore \frac{\mathrm{x}}{3}=1, \frac{\mathrm{y}}{3}=2, \frac{\mathrm{z}}{3}=3 \Rightarrow(\mathrm{x}, \mathrm{y}, \mathrm{z})=(3,6,9)
\)
Hence equation of required plane is
\(
\frac{x}{3}+\frac{y}{6}+\frac{z}{9}=1
\)
\(
\therefore \frac{\mathrm{x}}{3}=1, \frac{\mathrm{y}}{3}=2, \frac{\mathrm{z}}{3}=3 \Rightarrow(\mathrm{x}, \mathrm{y}, \mathrm{z})=(3,6,9)
\)
Hence equation of required plane is
\(
\frac{x}{3}+\frac{y}{6}+\frac{z}{9}=1
\)
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